Proving that $f$ is measurable with $f(x+y)= f(x)+f(y)$ then $f(x) =Ax$ for some $A\in\Bbb R$?

Question

How to show with given hints that $f$ is measurable with $f(x+y)= f(x)+f(y)$ implies that $f(x) =Ax$ for some $A\in\Bbb R$?

The following exercise is from an Analysis book by Eiolt-LieB, second edtion page 76:

Assume $f:\Bbb R\to \Bbb R$ is measurable such that $f(x+y)= f(x)+f(y).$ Prove that $f(x) =Ax$ for some $A\in\Bbb R.$

Hints:
a)Prove the result when $f$ is continuous.

b) Next: if $f$ is not continuous then, consider $f_\varepsilon(x) =\exp(if) \star j_\varepsilon(x)$. Where $ j_\varepsilon(x) =\varepsilon j(\varepsilon x)$ is a moliffier sequence. That is $j$ is smooth of compact support with $$\int_\Bbb R j(x) dx = 1.$$

I was able to solve the first question in the hint. But I don't how to use Hint(b). Would anyone help?

Also I would likt to know how to compute $$\lim_{\varepsilon \to 0 } f_\varepsilon(x)=?$$ I failed to use the Dominated Convergence Theorem.

This question has been asked in one of my previous questions: Let $g:\mathbb{R}\to\mathbb{R}$ be a measurable function such that $g(x+y) =g(x)+g(y).$ Then $g(x) = g(1)x$ .

But the solution there does not use the hints in this post.

Answer 1

Let's write $g \colon x \mapsto \exp\bigl(if(x)\bigr)$ and $g_{\varepsilon} = g \ast j_{\varepsilon}$. By standard theory, we know that $g_{\varepsilon}$ is continuous for every $\varepsilon > 0$. Also we know that for all $x,y\in \mathbb{R}$ we have

$$g_{\varepsilon}(x+y) = \int_{\mathbb{R}} g(x+y-t) j_{\varepsilon}(t)\,dt = \int_{\mathbb{R}} g(x)\cdot g(y-t)j_{\varepsilon}(t)\,dt = g(x)\cdot g_{\varepsilon}(y).$$

Thus, if $g_{\varepsilon}(y) \neq 0$ we have

$$g(x) = \frac{g_{\varepsilon}(x+y)}{g_{\varepsilon}(y)}$$

for all $x$, hence the continuity of $g$. This implies $g(x) = \exp(icx)$ for some $c\in \mathbb{R}$, and hence

$$f(x) = cx + h(x),$$

where $h \colon \mathbb{R} \to 2\pi \mathbb{Z} \subset \mathbb{R}$ is additive. That implies $h \equiv 0$. indeed assume $h(1)= 2\pi n$ with $n\neq0$ then since $h(x+y)=h(x)+h(y)$ we have $h(\frac{1}{n+1})= 2\pi \frac{n}{n+1}\in 2\pi\Bbb Z $ impossible.

So all that remains is to show that for suitable $y$ and $\varepsilon$ we have $g_{\varepsilon}(y) \neq 0$. That follows since $g_{\varepsilon} \xrightarrow{\varepsilon \downarrow 0} g$ in $L^1_{\text{loc}}(\mathbb{R})$, so there is a sequence $\varepsilon_k \to 0$ such that $g_{\varepsilon_k} \to g$ pointwise almost everywhere.